Plant Science 446/546. Final Examination May 16, 2002

Transcription

1 Plant Science 446/546 Final Examination May 16, 2002 Ag.Sci. Room :00am to 12:00 noon Name : Answer all 16 questions A total of 200 points are available A bonus question is available for an extra 10 points Questions are arranged into two sections Section I Selection/Novel Techniques Practical Applications Section II Basic concepts and Breeding Schemes Genetics/Selection Points available from each part of each question are shown in bold square parenthesis Try to be as brief and concise as possible Please write in a legible form Show any working/calculations Make sure that any additional paper used is attached to the questionnaire

2 2 SECTION I Selection/Novel Techniques Practical Applications Total of 100 points

3 3 1. Outline ALL the processes necessary to develop a new and improved cultivar using recombinant DNA technologies (plant transformation) [7 points]. Find a gene Clone the gene Find a transformation vector Produce a suitable construct Insert the construct Ensure that the gene in question is functional in the specific plant part Evaluate other agronomic characters Describe any difficulties that will need to be overcome in order for plant transformation techniques to become a major player in developing new crop cultivars [4 points]. o Find more valuable genes. o Transfer more than single genes 2. Describe the major differences in approach to selection in the early, intermediate and advanced stages of selection in a plant breeding program [7 points] and describe which experimental designs might be appropriate for each selection stage [3 points]. Early generation 1,000's of lines, no selection for quantitative traits, selection of single plots/plants, Un-replicated designs with replicated control plots. Intermediate generation 100's of lines, still variation for major characters, evaluation of quantitative traits, RCB, incomplete block designs. Advanced generation 10's of lines, larger plot trials, multiple locations throughout the target region, GxE determination and classification, Agronomic trials (i.e. vary N). RCB, incomplete block designs, factorial and split-plot designs.

4 4 3. List three processes that could utilize in vitro techniques in a plant breeding program. Increase clonal material Interspecific hybridization (embryo rescue) Protoplast fusion Plant transformation In vitro selection [3 points] List three problems that may be associated with in vitro techniques relating to plant breeding Unwanted genotypic selection Unwanted somatic mutation High cost Contamination/skilled staff [3 points] List two methods that could be utilized by plant breeders to increase the mutation events in plant breeding. o Chemical o Radiation [2 Points] Outline two factors that have to be considered when using mutagenesis techniques in a cultivar development scheme. o Safety o Unwanted deleterious mutation o Dose rate [2 points] 4. Why is an appropriate experimental design essential to plant breeding assessment trials? [3 points]. To obtain the 'best' error estimate to allow for statistical testing and other response prediction.

5 5 List 5 field factors that need to be considered when setting out plant breeding field trials Previous crop Fertility gradients Avoid, trees and buildings Guard rows Chemical history [5 points] Outline five difficulties that may be encountered when conducting evaluation trials off-station (i.e. regional testing): Distance from office Good cooperators/farmers Lack of on-site machinery Lack of storage and processing facilities Cost [5 points] List five uses that plant breeders could use glasshouses or growth chambers for: Pre-breeders seed increase Crossing In vitro to in vivo Disease or insect testing Rapid cycle to homozygosity [5 points] 5. A field trial is carried out whereby 400 F 4 progeny families are evaluated for yield potential at two different locations (Moscow and Colfax). The narrow-sense heritability for yield between the two sites is h = It is your intent to select the best 31 individual lines with general adaptability to the two environments and you wish to use independent culling to select the adapted cultivars. What percentage should you select at each location to ensure that 31 genotypes are selected at BOTH locations? [10 points]. Use inverse tetrachoric correlation table. As the heritability is 0.49, then the correlation coefficient r is lines from 400 are 7.75% so in the table we are looking for lines in the table. Value in table is 78, obtained by selecting 15% of lines from each site.

6 6 6. In 2000, 4,00 F 4 barley lines were evaluated for yield potential. The average yield of the complete population was 89.5 Bu/acre. The best lines were selected and their average yield potential was Bu/acre. A random sample of the initial population, along with the selected lines was grown in F 5 field yield trials in From these trials, the yield of the previously selected lines was Bu/acre and the yield of the random sample was 93.6 Bu/acre. From this determine the heritability for yield in barley and if the variance at F 5 is 35.4, determine the expected yield of the F 6 selected population, given that 10% of the best F 5 lines were retained (Hint: i = 10% selection) [10 points]. S = ; R = so h 2 = 0.49 Expected yield would be R, where R = ih 2 σ, (1.755 x 5.95 x 0.49) = Bu/acre doubled haploid lines of rice were produced and screened for molecular markers. It was found that these lines were polymorphic for 4 linked molecular markers, on chromosome #1. The 200 doubled haploid lines were evaluated for yield potential over four locations in Describe ALL the processes that would be involved in determining whether any quantitative trait loci (QTL) existed, and describe the process used to map position of these QTL s [10 points]. Screen 200 lines for 4 molecular markers Complete a one-way ANOVA to determine whether each marker is related to yield. Determine δ (delta) values associated with each marker Locate the QTL using multiple regressions. Describe two difficulties that might be encountered in utilizing QTL s in plant breeding selection [4 points]. Cost Differ with different crosses and different years and sites 8. Fully describe the procedure you might use to produce Registered seed for distribution to farmers from having an F 6 bulk selected population. Describe the procedure you would under take in each you of seed production [7 points]. Take 600 F 6 single plan selections, determine quality on seed from each plant. Only retain the ones with the 'best' stable quality and grow out as head-rows and check for variation as growing (Breeders' Seed).

7 7 At harvest, only retain the uniform plots and bulk harvest these. Plant these as Foundation seed and harvest this for Registered seed. In the Foundation and Registered seed stages, application must be made to the appropriate State representatives that will perform the mandated tests and field inspections to make sure seed meets the required standards. Describe what is meant by: DUS: Distinctness (different from others), Uniform (all the same as described), Stable (does not change over environments). VCU: Value for Cultivation and Use (must be agronomically adapted and be as good as or better than other cultivars. In relationship to plant breeding and cultivar release [4 points]. Describe three methods that could be used to obtain proprietary cultivar protection: PVP Patent Develop hybrids [6 points]

8 8 Section II Basic concepts and Breeding Schemes Genetics/Selection Total of 100 points

9 9 9. Three different genes have been identified for yellow stripe rust resistance in winter wheat. Each gene is positioned at different loci on the wheat genome. It is your intent to pyramid these genes (i.e. have all three gene in a single cultivar), describe the procedures you would use, using traditional breeding techniques, to develop a cultivar with all three gene, given that you start with three parents (P 1, P 2, and P 3 ) each with an different resistance gene [7 points]. Cross P 1 to P 2 screen F 2 for resistance. Self resistant lines and select lines which are homozygous for resistance gene(s). Test-cross selected lines for presence of 2 genes (i.e. ones that segregate 15:1, rather than 3:1 for resistance after testcross). Identify these plants and cross to P 3. Repeat the selection for homozygosity, and test cross to select plants with three genes The main thing is that it'll be difficult and require lots of crossing and test-crossing. Describe any advantages that you may have if there were molecular markers that existed for each resistance gene and explain the process you would follow given that you would select plants based on these molecular markers [3 points]. Using MM's it would be possible to make 3-way cross P 1 /P 2 //P 3 and screen F 2 plants, select lines with all three molecular markers, which should have a high probability of having all three genes. 10. List four features of a crop species that would merit hybrid cultivars to be developed: Means to produce hybrid seed High heterosis Cheap seed production Unable to produce uniform inbred crop. [4 points] Outline (using diagrams as necessary) a cytoplasmic male sterile system for producing hybrid seed involving a CMS genotype, a maintainer line, and fertility restoration line. Clearly label the cytoplasm and nuclear genes of importance in each line and the resulting hybrid. [6 points]. [A (CMS, rfrf) x A' (cms, rfrf)] x B(cms, RfRf), where CMS is cytoplasmic male sterile, but is overcome by dominant restorer genes (Rf).

10 Why is knowledge of plant evolution important to modern plant breeders? [4 points]. Many crop species have close relatives that evolved around the same center of origin. These can be invaluable sources of genes that can be introgressed into new cultivars. There has always been a question as to if ancient farmers cultivated crops or livestock first (or simultaneously), describe in your own words whether mankind cultivated crop or livestock first, and explain your answer [6 points]. I think animals, most likely horses that could be used to hunt and haul stuff from oasis to oasis. Farmers grew crops to feed their livestock throughout the year. 12. The following data is collected from a properly designed experiment of seed yield on segregating progeny and homozygous parents in a chickpea cross. Whereby the performance of both parents (P 1 and P 2 ), both backcross progeny (B 1 and B 2 ) and the F 3 progeny were evaluated. The data for family mean and within family variance are shown below. Design a suitable test and determine whether a simple additive/dominance model of inheritance is appropriate to explain the inheritance if seed yield in chickpea [12 points]. Family Sample size Mean yield Variance for yield P P B B F P 1 = m+a, P 2 = m-a, B 1 = m+1/2a+1/2d, B 2 = m-1/2a+1/2d, F 3 = m+1/4d 4F 3 B 1 B 2 - P 1 P 2 = T T = 2m + d m -1/2a 1/2d -1 +1/2a 1/2d m a m + a = 0, V(T) = 16VF 2 + VB 1 + VB 2 + VP 1 + VP 2 T=23.4, V(T) = 15,794 SE(T) = t-test = 23.4/125.6 with 125 d.f. = n.s. so the additive dominance model is adequate to explain the family variation. 13a. Two devoted friends have each been offered wonderful plant breeder jobs, one at the Drink mild beer barley breeding Company and the other at the Salty chips potato breeding Company. Compare any advantages and disadvantages that each friend will have to coupe with in setting the most suitable breeding scheme for each company [10 points].

11 11 The barley breeder will have the advantage of being able to drink beer (well maybe), while the potato breeder will get lots of chips. The barley breeder will most likely be developing inbred cultivars where he'll have difficult in selecting amongst segregating populations, or must attain near homozygosity before selecting. The advantage he'll have is that seed increase is relatively rapid. The potato breeder will have the advantage of genetically fixing any selection through vegetative propagation. However, potatoes are bulky are highly susceptible to disease through mother tubers. 13b. The friend has been given the task of selecting for only disease resistance combined with one of three dwarfing gene. All disease resistance in barley is controlled by single dominant alleles, while dwarfism is controlled by a single recessive allele. A fellow worker in the company breeds for only yield and quality, both of which are quantitatively inherited. Do you think that both breeders would adopt the same breeding scheme, and why? [12 points]. No I do not. Single gene characters are easier to select with small plots (or often a single plant), so pedigree selection may be appropritate for the disease program. The yield and quality program will require quantitative evaluation and hence a bulk based scheme may me more appropriate. 14. Breeding objectives are important when designing an appropriate breeding scheme. Briefly describe factors that would need to be considered is setting suitable breeding objectives [12 points]. People, from farmers, the first customer, to the final end-user, including all folks (i.e. seeds marketers, processors, etc.) in between. Politics Disease and pest spectrum Cultural farming practices (i.e. no-tillage systems, or organic systems) 15. Describe four ways that can be used to estimate heritability in plant breeding: By evaluating progeny Parent/off-spring regression Diallel analyses Response from selection [8 points] List three limitations of heritability in plant breeding:

12 12 Differ with cross Second order statistic Differ over years and sites [3 points] 16. A cross is made between two homozygous pea cultivars. One parent ( Rodman ) has round smooth seeds and long stems (RRLL); the other ( Prune Face ) has wrinkled seeds but with short stems (rrll). Round seed and long stems are both controlled by single dominant alleles (RR and LL, respectively). F 1 plants are crossed Prune Face and the progeny from this backcross grown out in the glasshouse. From 1,960 progeny examined the following phenotypes were observed: Round seed and long stems : 580 Round seed and short stems : 388 Wrinkled seed and long stems : 396 Wrinkled seed and short stems : 596 How many F 2 plants from the original cross (Rodman x Prune Face) would need to be grown to be 99% certain of obtaining at least one plant which was phenotypically short stemmed and with round smooth seeds? [13 points]. Recombination = 40% so Rl and rl recombinants occur at 0.2 frequency while RL and rl occur at 0.3 frequency. R_ll = RRll + Rrll = 0.16 or 16% of progeny Use Ln(1-p)/Ln(1-x) = Ln(0.01)/Ln(0.84) = Dam! I don t have a Ln calculator, but you get the idea. Bonus Questions [10 points] What will be the biggest challenges facing plant breeders in the next 20 years? Finding good graduate students. OR combining knowledge of quantitative genetics into molecular genetics. OR I don't care I retire in 9 years.